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<h1 class="heading"><a href="MATH-2023-OPDE.html"><span class="title">MATH 2023: Ordinary and Partial Differential Equations</span></a></h1>
<p class="byline">Xiaoyi Chen and Wei Zhang</p>
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<li class="link frontmatter"><a href="meta_frontmatter.html" data-scroll="meta_frontmatter" class="internal"><span class="title">Front Matter</span></a></li>
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<a href="ch_first.html" data-scroll="ch_first" class="internal"><span class="codenumber">1</span> <span class="title">Introduction</span></a><ul>
<li><a href="sec_1-intro.html" data-scroll="sec_1-intro" class="internal">Classification of Differential Equations</a></li>
<li><a href="sec_2-intro.html" data-scroll="sec_2-intro" class="internal">Linear and Nonlinear Equation</a></li>
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<a href="ch_second.html" data-scroll="ch_second" class="internal"><span class="codenumber">2</span> <span class="title">First Order Ordinary Differential Equations</span></a><ul>
<li><a href="sec2_1.html" data-scroll="sec2_1" class="internal">Linear Equations</a></li>
<li><a href="sec2_2.html" data-scroll="sec2_2" class="internal">Further Discussion of Linear Equations (For reading only)</a></li>
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<a href="ch_third.html" data-scroll="ch_third" class="internal"><span class="codenumber">3</span> <span class="title">third Order Linear Equations</span></a><ul>
<li><a href="sec3_1.html" data-scroll="sec3_1" class="internal">Homogeneous equations with constant coefficient</a></li>
<li><a href="sec3_2.html" data-scroll="sec3_2" class="internal">Fundamental Solutions of Linear Homogeneous Equations</a></li>
<li><a href="sec3_3.html" data-scroll="sec3_3" class="internal">Linear Independence and Wronskian</a></li>
<li><a href="sec3_4.html" data-scroll="sec3_4" class="internal">Complex roots of the characteristic equations</a></li>
<li><a href="sec3_5.html" data-scroll="sec3_5" class="active">Repeated Roots: Reduction of Order</a></li>
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<li><a href="sec3_7.html" data-scroll="sec3_7" class="internal">Variation of Parameters</a></li>
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<li><a href="sec4_1.html" data-scroll="sec4_1" class="internal">General Theory of the <span class="process-math">\(n\)</span>-th Order Linear Equations</a></li>
<li><a href="sec4_2.html" data-scroll="sec4_2" class="internal">Homogeneous Equations with Constant Coefficients</a></li>
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<li><a href="sec4_4.html" data-scroll="sec4_4" class="internal">The Method of Variation of Parameters</a></li>
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<li><a href="sec5_1.html" data-scroll="sec5_1" class="internal">Brief Review on Power Series</a></li>
<li><a href="sec5_2.html" data-scroll="sec5_2" class="internal">Introduction</a></li>
<li><a href="sec5_3.html" data-scroll="sec5_3" class="internal">Series Solutions Near an Ordinary Point</a></li>
<li><a href="sec5_4.html" data-scroll="sec5_4" class="internal">Euler’s Equation</a></li>
<li><a href="sec5_5.html" data-scroll="sec5_5" class="internal">Series Solution near a Regular Singular Point</a></li>
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<a href="ch_six.html" data-scroll="ch_six" class="internal"><span class="codenumber">6</span> <span class="title">System of First Order Linear Equations</span></a><ul>
<li><a href="sec6_1.html" data-scroll="sec6_1" class="internal">Introduction <span class="process-math">\(\&amp;\)</span> Basic Theory</a></li>
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<a href="ch_seven.html" data-scroll="ch_seven" class="internal"><span class="codenumber">7</span> <span class="title">Partial Differential Equations</span></a><ul>
<li><a href="sec7_1.html" data-scroll="sec7_1" class="internal">Two-Point Boundary Value Problems</a></li>
<li><a href="sec7_2.html" data-scroll="sec7_2" class="internal">Eigenvalue Problems</a></li>
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<li><a href="sec7_5.html" data-scroll="sec7_5" class="internal">Even and Odd Functions</a></li>
<li><a href="sec7_6.html" data-scroll="sec7_6" class="internal">Introduction to Partial Differential Equations</a></li>
<li><a href="sec7_7.html" data-scroll="sec7_7" class="internal">1D Heat Equation; Solutions by Separation of Variable and Fourier Series</a></li>
<li><a href="sec7_8.html" data-scroll="sec7_8" class="internal">Other Heat Conduction Problems</a></li>
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<a href="ch_eight.html" data-scroll="ch_eight" class="internal"><span class="codenumber">8</span> <span class="title">Laplace transform</span></a><ul>
<li><a href="sec8_1.html" data-scroll="sec8_1" class="internal">What are Laplace Transforms, and Why?</a></li>
<li><a href="sec8_2.html" data-scroll="sec8_2" class="internal">Finding Laplace Transforms</a></li>
<li><a href="sec8_3.html" data-scroll="sec8_3" class="internal">Finding inverse transforms using partial fractions</a></li>
<li><a href="sec8_4.html" data-scroll="sec8_4" class="internal">Solving ODEs and ODE Systems</a></li>
<li><a href="sec8_5.html" data-scroll="sec8_5" class="internal">Step input and Impulse problems</a></li>
<li><a href="sec8_6.html" data-scroll="sec8_6" class="internal">Laplace transform for PDE (heat equation)</a></li>
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<li class="link"><a href="solutions-1.html" data-scroll="solutions-1" class="internal"><span class="codenumber">A</span> <span class="title">Selected Hints</span></a></li>
<li class="link"><a href="solutions-2.html" data-scroll="solutions-2" class="internal"><span class="codenumber">B</span> <span class="title">Selected Solutions</span></a></li>
<li class="link"><a href="appendix-1.html" data-scroll="appendix-1" class="internal"><span class="codenumber">C</span> <span class="title">List of Symbols</span></a></li>
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<main class="main"><div id="content" class="pretext-content"><section class="section" id="sec3_5"><h2 class="heading hide-type">
<span class="type">Section</span> <span class="codenumber">3.5</span> <span class="title">Repeated Roots: Reduction of Order</span>
</h2>
<p id="p-102">Consider</p>
<div class="displaymath process-math" data-contains-math-knowls="">
\begin{equation*}
y^{\prime \prime}+b y^{\prime}+c y=0.
\end{equation*}
</div>
<p class="continuation">The characteristic equation:</p>
<div class="displaymath process-math" data-contains-math-knowls="">
\begin{equation*}
r^2+b r+c=0.
\end{equation*}
</div>
<p class="continuation">We consider the case <span class="process-math">\(b^2-4 c=0\text{,}\)</span> i.e., there are two repeated roots: <span class="process-math">\(r_1=r_2=-\frac{b}{2}\text{.}\)</span> One solution is <span class="process-math">\(y_1=e^{-\frac{b}{2} x}\text{.}\)</span><dfn class="terminology">How to find another linear independent solution? (Reduction of order)</dfn></p>
<p id="p-103">We know that <span class="process-math">\(C_1 y_1\)</span> satisfies the ODE. We replace <span class="process-math">\(C_1\)</span> by a function of <span class="process-math">\(x\text{,}\)</span> say <span class="process-math">\(v(x)\text{.}\)</span> Then, we try to determine <span class="process-math">\(v(x)\)</span> to make <span class="process-math">\(y_2=v(x) y_1\)</span> be a solution which is also independent of <span class="process-math">\(y_1\text{.}\)</span> Substituting <span class="process-math">\(y_2=v(x) y_1\)</span> into the ODE, we have</p>
<div class="displaymath process-math" data-contains-math-knowls="">
\begin{equation*}
\begin{aligned}
&amp;\quad ~ v^{\prime \prime} y_1+2 v^{\prime} y_1^{\prime}+v y_1^{\prime \prime}+b (v^{\prime} y_1+v y_1^{\prime})+c v y_1=0\\
&amp;\to~v (y_1^{\prime \prime}+b y_1^{\prime}+ c y_1)+v^{\prime \prime} y_1+2 v^{\prime} y_1^{\prime}+b v^{\prime}y_1=0
\\
&amp;\to~ 0+v^{\prime \prime} y_1+v^{\prime} (2 y_1^{\prime}+b y_1)=0\\
&amp;\to~v^{\prime \prime} e^{-\frac{b}{2}x}+v^{\prime} \cdot 0=0\\
&amp;\to~v^{\prime \prime}=0\\
&amp;\to~v(x)=d_1 x+d_2,
\end{aligned}
\end{equation*}
</div>
<p class="continuation">where <span class="process-math">\(d_1, d_2\)</span> are arbitrary constants.</p>
<p id="p-104">We only need one form of <span class="process-math">\(v(x)\)</span> as long as <span class="process-math">\(v(x) y_1\)</span> is a solution linearly independent of <span class="process-math">\(y_1\text{.}\)</span> Setting <span class="process-math">\(d_1=1\)</span> and <span class="process-math">\(d_2=0\text{,}\)</span> we have</p>
<div class="displaymath process-math" data-contains-math-knowls="">
\begin{equation*}
y_2=x y_1=x e^{-\frac{b}{2} x}.
\end{equation*}
</div>
<p class="continuation">Further,</p>
<div class="displaymath process-math" data-contains-math-knowls="">
\begin{equation*}
W=\left| \begin{array}{ll}
y_1 &amp; y_2\\
y_1^{\prime} &amp; y_2^{\prime}
\end{array}\right|=
\left| \begin{array}{lll}
e^{-\frac{b}{2}x} &amp; \quad &amp; x e^{-\frac{b}{2}x}\\
-\frac{b}{2} e^{-\frac{b}{2} x} &amp; \quad &amp; e^{-\frac{b}{2}x}-\frac{b x}{2} e^{-\frac{b}{2} x}
\end{array} \right|=e^{-bx} \neq 0.
\end{equation*}
</div>
<p class="continuation">This means <span class="process-math">\(y_1\)</span> and <span class="process-math">\(y_2\)</span> are linear independent. So the general solution is</p>
<div class="displaymath process-math" data-contains-math-knowls="">
\begin{equation*}
y=C_1 y_1+C_2 y_2=C_1 e^{-\frac{b}{2}x}+C_2 x  e^{-\frac{b}{2}x}.
\end{equation*}
</div>
<p id="p-105"><dfn class="terminology">Example 1</dfn> Find the general solution of the following ODE</p>
<div class="displaymath process-math" data-contains-math-knowls="">
\begin{equation*}
4 y^{\prime \prime}+12 y^{\prime}+9 y=0.
\end{equation*}
</div>
<p id="p-106"><dfn class="terminology">Solution</dfn> The characteristic equation is</p>
<div class="displaymath process-math" data-contains-math-knowls="">
\begin{equation*}
4 r^2+12 r+9=0~\to~(2 r+3)^2=0~\to~r_1=-\frac{3}{2}=r_2.
\end{equation*}
</div>
<p class="continuation">Thus one solution is</p>
<div class="displaymath process-math" data-contains-math-knowls="">
\begin{equation*}
y_1=e^{-\frac{3}{2} x}.
\end{equation*}
</div>
<p class="continuation">The second solution is</p>
<div class="displaymath process-math" data-contains-math-knowls="">
\begin{equation*}
y_2=x y_1=x e^{-\frac{3}{2}x} .
\end{equation*}
</div>
<p class="continuation">The general solution is</p>
<div class="displaymath process-math" data-contains-math-knowls="">
\begin{equation*}
y=C_1 e^{-\frac{3}{2} x}+C_2 x e^{-\frac{3}{2} x}.
\end{equation*}
</div>
<p id="p-107"><dfn class="terminology">Example 2</dfn> Find the solution of the following initial value problem</p>
<div class="displaymath process-math" data-contains-math-knowls="">
\begin{equation*}
y^{\prime \prime}-6 y^{\prime}+9y=0, \quad y(0)=0,\quad y^{\prime}(0)=2.
\end{equation*}
</div>
<p id="p-108"><dfn class="terminology">Solution</dfn> The characteristic equation is</p>
<div class="displaymath process-math" data-contains-math-knowls="">
\begin{equation*}
r^2-6r+9=0~\to~(r-3)^2=0~\to~r_1=r_2=3.
\end{equation*}
</div>
<p class="continuation">One solution is</p>
<div class="displaymath process-math" data-contains-math-knowls="">
\begin{equation*}
y_1=e^{3 x}.
\end{equation*}
</div>
<p class="continuation">The second solution is</p>
<div class="displaymath process-math" data-contains-math-knowls="">
\begin{equation*}
y_2=x y_1=x e^{3x}.
\end{equation*}
</div>
<p class="continuation">The general solution is</p>
<div class="displaymath process-math" data-contains-math-knowls="">
\begin{equation*}
y=C_1 e^{3x}+C_2 x e^{3x}.
\end{equation*}
</div>
<p class="continuation">The initial conditions yield</p>
<div class="displaymath process-math" data-contains-math-knowls="">
\begin{equation*}
\begin{aligned}
&amp;y(0)=0:\quad 0=C_1 ~\to~y=C_2 x e^{3x}\\
&amp;y^{\prime}(0)=2 ~\to~2=C_2.
\end{aligned}
\end{equation*}
</div>
<p class="continuation">The solution is</p>
<div class="displaymath process-math" data-contains-math-knowls="">
\begin{equation*}
y=2 x e^{3x}.
\end{equation*}
</div>
<p id="p-109"><dfn class="terminology">Reduction of Order (For reading only)</dfn>Suppose that <span class="process-math">\(y_1(x)\)</span> is a solution of</p>
<div class="displaymath process-math" data-contains-math-knowls="">
\begin{equation*}
y^{\prime \prime}+p(x) y^{\prime}+q(x) y=0.
\end{equation*}
</div>
<p class="continuation">To find another solution, we let <span class="process-math">\(y_2=v(x) y_1\text{.}\)</span> Substituting it into the ODE,</p>
<div class="displaymath process-math" data-contains-math-knowls="">
\begin{equation*}
\begin{aligned}
&amp;v^{\prime \prime} y_1+2 v^{\prime} y_1^{\prime}+v y_1^{\prime \prime}+p(x) (v^{\prime} y_1+v y_1^{\prime})+q(x) v y_1=0\\
&amp; v [y_1^{\prime \prime}+p(x)y_1^{\prime}+q(x) y_1]+v^{\prime \prime} y_1+v^{\prime}[2 y_1^{\prime}+p(x) y_1]=0\\
&amp;\qquad \qquad \qquad \qquad \quad 0+v^{\prime \prime} y_1+v^{\prime}[2 y_1^{\prime}+p(x) y_1]=0.
\end{aligned}
\end{equation*}
</div>
<p class="continuation">Setting <span class="process-math">\(u=\frac{\textrm{d} v}{\textrm{d} x}\text{,}\)</span> one has</p>
<div class="displaymath process-math" data-contains-math-knowls="">
\begin{equation*}
y_1 \frac{\textrm{d} u}{\textrm{d} x}+[2 y_1^{\prime}+p(x) y_1] u=0.
\end{equation*}
</div>
<p class="continuation">This is a first order ODE for <span class="process-math">\(u\text{,}\)</span> you can solve it by using the integrating factor.</p>
<p id="p-110"><dfn class="terminology">Example 1</dfn> Suppose that <span class="process-math">\(y_1(x)=\sin x^2\)</span> is a solution of</p>
<div class="displaymath process-math" data-contains-math-knowls="">
\begin{equation*}
x y^{\prime \prime}-y^{\prime}+4 x^3 y=0,\quad x&gt;0.
\end{equation*}
</div>
<p class="continuation">Find the general solution.</p>
<p id="p-111"><dfn class="terminology">Solution</dfn> Seek a solution of the form</p>
<div class="displaymath process-math" data-contains-math-knowls="">
\begin{equation*}
y_2=v(x) y_1=v(x) \sin x^2.
\end{equation*}
</div>
<p class="continuation">Substituting it into the ODE, one has</p>
<div class="displaymath process-math" data-contains-math-knowls="">
\begin{equation*}
\begin{aligned}
&amp; x [v^{\prime \prime} y_1+2 v^{\prime} y_1^{\prime}+v y_1^{\prime \prime}]-(v^{\prime} y_1+v y_1^{\prime}) +4 x^3 v y_1=0\\
&amp;\to~ v [ x y_1^{\prime \prime} -y_1^{\prime}+4 x^3 y_1]+x y_1 v^{\prime \prime}+v^{\prime} [2 x y_1^{\prime}-y_1]=0\\
&amp;\to~\qquad \qquad \qquad \quad 0+x y_1 v^{\prime \prime}+v^{\prime} [2 x y_1^{\prime}-y_1]=0.
\end{aligned}
\end{equation*}
</div>
<p class="continuation">Let <span class="process-math">\(u=\frac{\textrm{d} v}{\textrm{d} x}\text{,}\)</span> we have</p>
<div class="displaymath process-math" data-contains-math-knowls="">
\begin{equation*}
\begin{aligned}
&amp;x \sin x^2 \frac{\textrm{d} u}{\textrm{d} x}+[4 x^2 \cos x^2-\sin x^2 ] u=0\\
&amp;\to~\frac{\textrm{d} u}{u}=-\frac{4 x^2 \cos x^2-\sin x^2}{x \sin x^2}  \textrm{d} x=-\frac{4 x \cos x^2}{\sin x^2} \textrm{d} x+\frac{1}{x} \textrm{d} x\\
&amp;\ln |u|=-\int \frac{2 \textrm{d} \sin x^2}{\sin x^2}+\ln |x|+C_1=-2 \ln |\sin x^2|+\ln |x|+C_1\\
&amp;\to~u=C_2 \frac{x}{(\sin x^2)^2}.
\end{aligned}
\end{equation*}
</div>
<div class="displaymath process-math" data-contains-math-knowls="" id="p-112">
\begin{equation*}
\frac{\textrm{d} v}{\textrm{d} x}=C_2 \frac{x}{(\sin x^2)^2}~\to~v=C_2 \int \frac{x}{(\sin x^2)^2} \textrm{d} x+C_3.
\end{equation*}
</div>
<p class="continuation">Letting <span class="process-math">\(t=x^2\text{,}\)</span> one has <span class="process-math">\(\textrm{d} t=2x \textrm{d} x\)</span> and the above solution to <span class="process-math">\(v\)</span> becomes</p>
<div class="displaymath process-math" data-contains-math-knowls="">
\begin{equation*}
v=C_2 \int \frac{\frac{1}{2}}{(\sin t)^2}    \textrm{d}t+C_3=-\frac{C_2}{2} \cot t+C_3=-\frac{C_2}{2} \cot x^2+C_3.
\end{equation*}
</div>
<p class="continuation">Choose <span class="process-math">\(C_2=-2, C_3=0\text{,}\)</span> we have the second solution to be</p>
<div class="displaymath process-math" data-contains-math-knowls="">
\begin{equation*}
y_2=v y_1=\cot x^2 \sin x^2=\cos x^2.
\end{equation*}
</div>
<p class="continuation">Check</p>
<div class="displaymath process-math" data-contains-math-knowls="">
\begin{equation*}
W(y_1, y_2)=\left|
\begin{array}{ll}
y_1 &amp; y_2\\
y_1^{\prime} &amp; y_2^{\prime}
\end{array}
\right|
=
\left|
\begin{array}{ll}
\sin x^2 &amp; \cos x^2\\
2 x \cos x^2 &amp; -2 x \sin x^2
\end{array}
\right|=-2 x \neq 0.
\end{equation*}
</div>
<p class="continuation">Thus, the general solution is</p>
<div class="displaymath process-math" data-contains-math-knowls="">
\begin{equation*}
y=C_1 \sin x^2+C_2 \cos x^2.
\end{equation*}
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